Whirling of shafts — a disc fixed at mid-span has CG eccentricity e = 2 mm relative to the shaft axis. The lateral natural frequency is 10 rad/s. If the shaft rotates at 300 r.p.m., the steady whirl radius will be approximately:

Introduction / Context: An unbalanced rotor (disc with eccentric center of gravity) mounted on a flexible shaft will whirl at a radius that depends on running speed relative to the lateral natural frequency. This is the classical Jeffcott (Laval) rotor model.

Given Data / Assumptions:

• Eccentricity e = 2 mm = 0.002 m.
• Undamped lateral natural frequency ω_n = 10 rad/s.
• Running speed 300 r.p.m. ⇒ ω = 2π×300/60 = 10π ≈ 31.416 rad/s.
• Neglect damping for the amplitude estimate.

Concept / Approach: For an undamped Jeffcott rotor, steady whirl amplitude X equals X = e·ω²/|ω_n² − ω²|. Since ω > ω_n here, use X = e·ω²/(ω² − ω_n²).

Step-by-Step Solution:

Verification / Alternative check: Limiting cases: as ω→ω_n, amplitude grows large (resonance); as ω ≫ ω_n, X→e (approaches the geometric eccentricity), which aligns with the computed value being slightly above 2 mm.

Why Other Options Are Wrong:
2 mm — would be the asymptote as ω ≫ ω_n, but ω is only ~3.14× ω_n.
2.50 mm and 3.0 mm — overestimate the dynamic amplification at this speed ratio.

Common Pitfalls: Using (ω_n² − ω²) in the denominator for ω > ω_n; forgetting to convert r.p.m. to rad/s.

Final Answer: 2.22 mm.